# Nested open sets?

1. Mar 18, 2012

### cragar

1. The problem statement, all variables and given/known data
Give an example of an infinite collection of nested open sets.
$o_1 \supseteq o_2 \supseteq o_3 \supseteq o_4 .....$
Whose intersection $\bigcap_{n=1}^{ \infty} O_n$ is
closed and non empty.
2. Relevant equations
A set $O \subseteq \mathbb{R}$ is open if for all points, $a \in O$
there exists an $\epsilon$ neighborhood $V_{\epsilon}(a) \subseteq O$
3. The attempt at a solution
It seems like if we started with the open interval (0,1) and then took a smaller interval that was nested inside the original interval, and then just kept doing this until we enclosed one point in the interval.

2. Mar 18, 2012

### micromass

Staff Emeritus
Yes, that's the idea.

Can you make this rigorous??

3. Mar 18, 2012

### cragar

Can I just take the middle one-third of the set.
so after n operations i will have $\frac{1}{3^n}$

4. Mar 18, 2012

### jgens

I cannot really tell where you are going with that last response. It might be a little easier if you consider intervals of the form $(-\frac{1}{n},\frac{1}{n})$.

5. Mar 18, 2012

### micromass

Staff Emeritus
What do you mean?? 1/3^n is just a number.

Just find sets that get smaller and smaller each time.

6. Mar 18, 2012

### cragar

ok so like jgens said use $( \frac{-1}{n} , \frac{1}{n})$
And then eventually after n goes to infinity I will have 0 as my enclosed point.
so if I make an $\epsilon$ radius around 0 i will contain points inside of O the original set. Would the set zero it self be closed be cause if we make
an $\epsilon$ radius around 0 it wont contain elements that are in the set zero itself.

7. Mar 18, 2012

### HallsofIvy

Staff Emeritus
Or, if you really want $1/3^n$, you could use $O_n= (-1/3^n, 1/3^n)$.

8. Mar 18, 2012

### cragar

ok, thanks everyone for the help